Volume of a solid region enclosed by elliptic cylinder using triple integral

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Find the volume of a solid limited laterally by an elliptical cylinder $$\frac{(x-2)^2}{4}+\frac{(y-1)^2}{9}=1$$ and by the planes $z+x=5$ and $z+x=6$.

I tried doing by cylindrical coordinates and Cartesian, but I could not get to the answer, which is $6\pi$. Can anyone give me some ideas on how to proceed?

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Please use change of variable $x = 2 + 2r \cos \theta, y = 1 + 3r \sin\theta$

Your ellipse transforms to a circle of radius $1$ and so we have $0 \leq r \leq 1, 0 \leq \theta \leq 2 \pi$.

and the limits of $z$ will be given by planes

$3 - 2 r \cos \theta \leq z \leq 4 - 2r \cos \theta$.

Jacobian for the change of variables,

$|J| = 6r$

Can you take it from here?